An observer is 28 m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 26 m horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that the observer's line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of 6 m/s, what is the rate of change of the angle of elevation when the elevator is 15 m above the ground? When the elevator is 54 m above the ground?

Question

An observer is 28 m above the ground floor of a large hotel atrium looking at a​ glass-enclosed elevator shaft that is 26 m horizontally from the observer​ (see figure). The angle of elevation of the elevator is the angle that the​ observer's line of sight makes with the horizontal​ (it may be positive or​ negative). Assuming that the elevator rises at a rate of 6 m/s​, what is the rate of change of the angle of elevation when the elevator is 15 m above the​ ground? When the elevator is 54 m above the​ ground?

oobleck · Accepted Answer

when the elevator is x meters up, then we have the angle of elevation θ is
tanθ = (x-28)/26 for 0<= x <= (some max height)
sec^2θ dθ/dt = 1/26 dx/dt
so, when x=15,
tanθ = -13/26 = -1/2
sec^2θ = 1 + tan^2θ = 5/4
and we have
5/4 dθ/dt = 1/26 * 6
dθ/dt = 12/65 rad/s

Noah · Answer

When the elevator is 54 m above the​ ground?

oobleck · Answer

c'mon dude - use x=54 you gotta do some of the work! Post your work if you get stuck.